Question

1. of 5

time (hours) temperature (°f)
0 104.8
0.5 103.6
1 102.9
1.5 102.6
2 102.3
2.5 102.0
3 101.7
3.5 101.2
4 99.4
linear equation:
r =
type of correlation:
what is the prediction for 11 hours?

Explanation:

Step1: Recall linear - regression formula

The general form of a linear regression equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. To find $m$ and $b$, we use the following formulas:
\[m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}\]
\[b=\frac{\sum_{i = 1}^{n}y_i - m\sum_{i = 1}^{n}x_i}{n}\]
Let $x$ be the time (hours) and $y$ be the temperature ($^{\circ}F$). We have $n = 9$ data - points: $(x_1,y_1)=(0,104.8),(x_2,y_2)=(0.5,103.6),\cdots,(x_9,y_9)=(4,99.4)$.
First, calculate the necessary sums:
\(\sum_{i = 1}^{9}x_i=0 + 0.5+1+1.5+2+2.5+3+3.5+4 = 18\)
\(\sum_{i = 1}^{9}y_i=104.8 + 103.6+102.9+102.6+102.3+102.0+101.7+101.2+99.4 = 910.5\)
\(\sum_{i = 1}^{9}x_i^{2}=0^{2}+0.5^{2}+1^{2}+1.5^{2}+2^{2}+2.5^{2}+3^{2}+3.5^{2}+4^{2}=0 + 0.25+1+2.25+4+6.25+9+12.25+16 = 51\)
\(\sum_{i = 1}^{9}x_iy_i=0\times104.8+0.5\times103.6+1\times102.9+1.5\times102.6+2\times102.3+2.5\times102.0+3\times101.7+3.5\times101.2+4\times99.4\)
\(=0 + 51.8+102.9+153.9+204.6+255+305.1+354.2+397.6 = 1825.1\)

Step2: Calculate the slope $m$

\[m=\frac{9\times1825.1-18\times910.5}{9\times51-(18)^{2}}=\frac{16425.9 - 16389}{459 - 324}=\frac{36.9}{135}\approx - 0.2733\]

Step3: Calculate the y - intercept $b$

\[b=\frac{910.5-(-0.2733)\times18}{9}=\frac{910.5 + 4.9194}{9}=\frac{915.4194}{9}\approx101.7133\]
So the linear equation is $y=-0.2733x + 101.7133$.

Step4: Calculate the correlation coefficient $r$

The formula for the correlation coefficient $r$ is:
\[r=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{\sqrt{(n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2})(n\sum_{i = 1}^{n}y_i^{2}-(\sum_{i = 1}^{n}y_i)^{2})}}\]
First, calculate \(\sum_{i = 1}^{9}y_i^{2}=104.8^{2}+103.6^{2}+102.9^{2}+102.6^{2}+102.3^{2}+102.0^{2}+101.7^{2}+101.2^{2}+99.4^{2}\)
\(=10983.04+10732.96+10588.41+10526.76+10465.29+10404+10342.89+10241.44+9880.36 = 94165.14\)
\[r=\frac{9\times1825.1-18\times910.5}{\sqrt{(9\times51-(18)^{2})(9\times94165.14-(910.5)^{2})}}\]
\[=\frac{36.9}{\sqrt{135\times(847486.26 - 829010.25)}}=\frac{36.9}{\sqrt{135\times18476.01}}=\frac{36.9}{\sqrt{2494261.35}}\approx - 0.74\]
The type of correlation is negative since $r\lt0$.

Step5: Make a prediction for $x = 11$

Substitute $x = 11$ into the linear equation $y=-0.2733x + 101.7133$.
\[y=-0.2733\times11 + 101.7133=-3.0063+101.7133 = 98.707\approx98.7\]

Answer:

Linear Equation: $y=-0.2733x + 101.7133$
$r=-0.74$
type of correlation: negative
What is the prediction for 11 hours? $98.7$